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NIMCET Previous Year Questions (PYQs)

NIMCET Progressions PYQ

Qus : 1NIMCET PYQ 2019

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function

in the interval [-2,3] and the difference between the first two terms is f'(0). Then the common ratio of GP is

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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Qus : 2NIMCET PYQ 2019

$a, b, c$ are in GP and

$log a - log 2b$ ,

$log 2b - log 3c$ and

$log 3c - log a$ are in AP, then

$a, b, c$ are the lengths of the sides of a triangle which is

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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Qus : 3NIMCET PYQ 2019

Not Available

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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Qus : 4NIMCET PYQ 2019

$a, a, a_2, ., a_{2n-1},b$ are in AP,

$a, b_1, b_2,...b_{2n-1}, b$ are in GP and

$a, c_1, c_2,... c_{2n-1}, b$ are in HP, where a, b are positive, then the equation

$a_n x^2-b_n+c_n$ has its roots

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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Qus : 5NIMCET PYQ 2024

$x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then

${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =

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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Given:

$x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6}$

Step 1: Write in powers of $a = 2^{1/6}$

$x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1}$

Step 2: Use identity $a^6 = 2 \Rightarrow a^5 = \frac{2}{a}$

$x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24}$

Step 3: Final calculation

$a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16}$

✅ Final Answer: $\boxed{16}$

Qus : 6NIMCET PYQ 2024

The number of solutions of

${5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25$ for

$x\in(-\mathrm{\pi},\mathrm{\pi})$ is

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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Step 1: Recognize the series

This is a geometric series with first term $a = 1$ , common ratio $r = |\sin x| \in [0,1]$ , so: $\text{Sum} = \frac{1}{1 - |\sin x|}$

Step 2: Rewrite the equation

$5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2$

Equating exponents: $\frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2}$

Step 3: Solve for $x \in (-\pi, \pi)$

We want all $x \in (-\pi, \pi)$ such that $|\sin x| = \frac{1}{2}$

So $\sin x = \pm \frac{1}{2}$ . Within $(-\pi, \pi)$ , the values of $x$ satisfying this are:

$x = \frac{\pi}{6}$
$x = \frac{5\pi}{6}$
$x = -\frac{\pi}{6}$
$x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Qus : 7NIMCET PYQ 2022

Which term of the series

$\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is

$\frac{\sqrt{5}}{13}$ ?

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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

Qus : 8NIMCET PYQ 2022

In a Harmonic Progression,

$p^{th}$ term is

$q$ and the

$q^{th}$ term is

$p$ . Then

$pq^{th}$ term is

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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

Qus : 9NIMCET PYQ 2024

If one AM (Arithmetic mean) 'a' and two GM's (Geometric means) p and q be inserted between any two positive numbers, the value of p^3+q^3 is

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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Problem:

If one Arithmetic Mean (AM) $a$ and two Geometric Means $p$ and $q$ are inserted between any two positive numbers, find the value of: $p^3 + q^3$

Given:

Let two positive numbers be $A$ and $B$ .
One AM: $a = \frac{A + B}{2}$
Two GMs inserted: so the four terms in G.P. are: $A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B$

Now calculate:

$pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB$
$p^3 = A^2B, \quad q^3 = AB^2$
$p^3 + q^3 = A^2B + AB^2 = AB(A + B)$

Also,

$2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B)$

✅ Therefore,

$\boxed{p^3 + q^3 = 2apq}$

Qus : 10NIMCET PYQ 2016

The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is

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NIMCET Previous Year PYQNIMCET NIMCET 2016 PYQ

Solution

Qus : 11NIMCET PYQ 2018

Sum to infinity of a geometric is twice the sum of the first two terms. Then what are the possible values of common ratio?

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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

Solution

Qus : 12NIMCET PYQ 2018

Suppose that m and n are fixed numbers such that the mth term am is equal to n and nth term an is equal to m, (m≠n), the the value of (m+n)th term is

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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

Solution

Qus : 13NIMCET PYQ 2017

The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation 2A+G² = 27, then the two numbers are

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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ

Solution

Qus : 14NIMCET PYQ 2017

Three positive number whose sum is 21 are in arithmetic progression. If 2, 2, 14 are added to them respectively then resulting numbers are in geometric progression. Then which of the following is not among the three numbers?

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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ

Solution

Let the three terms in A.P. be a – d, a, a + d.

given that a – d + a + a + d = 21

a = 7

then the three term in A.P. are 7 – d, 7, 7 + d

According to given condition 9 – d, 9, 21 + d are in G.P.

(9)² = (9 – d) (21 + d)

81 = 189 + 9d – 21d – d²

81 = 189 – 12d – d²

d² + 12d – 108 = 0

d(d + 18) – 6 (d + 18) = 0

(d – 6) (d + 18) = 0

We get, d = 6, –18

Putting d = 6 in the term 7 – d, 7, 7 + d we get 1, 7, 13.

Qus : 15NIMCET PYQ 2021

$H_1,H_2,\ldots,H_n$ are n harmonic means between a and b

$(b\ne a)$ ;,then

$\frac{{{H}}_n+a}{{{H}}_n-a}+\frac{{{H}}_n+b}{{{H}}_n-b}$

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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

Qus : 16NIMCET PYQ 2018

are positive real numbers whose product is a fixed number c, then the minimum of

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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

Solution

Qus : 17NIMCET PYQ 2021

The four geometric means between 2 and 64 are

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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

Qus : 18NIMCET PYQ 2015

If a, b, c are in geometric progression, then

$log_{ax}^{a}, log_{bx}^{a}$ and

$log_{cx}^{a}$ are in

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NIMCET Previous Year PYQNIMCET NIMCET 2015 PYQ

Solution

Qus : 19NIMCET PYQ 2015

The value of the sum

$\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}$ is

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NIMCET Previous Year PYQNIMCET NIMCET 2015 PYQ

Solution

Qus : 20NIMCET PYQ 2020

An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Then what is the common difference?

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NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

Solution

Qus : 21NIMCET PYQ 2023

The sum of infinite terms of decreasing GP is equal to the greatest value of the function

$f(x) = x^3 + 3x – 9$ in the interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is

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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution

? GP and Function Relation

Given: $f(x) = x^3 + 3x - 9$

The sum of infinite GP = max value of $f(x)$ on [−2, 3]

The difference between first two terms = $f'(0)$

Step 1: $f(x)$ is increasing ⇒ Max at $x = 3$

$f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27$

Step 2: $f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3$

⇒ $a(1 - r) = 3$

Step 3: Solve:
$a = 27(1 - r)$
$\Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3}$

✅ Final Answer: $r = \frac{2}{3}$

Qus : 22NIMCET PYQ 2023

If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?

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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution

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NIMCET Previous Year Questions (PYQs)

NIMCET Progressions PYQ

Solution

Solution

Solution

Solution

Solution

Given:

Step 1: Write in powers of a=21/6 a = 2^{1/6}

Step 2: Use identity a6=2⇒a5=2a a^6 = 2 \Rightarrow a^5 = \frac{2}{a}

Step 3: Final calculation

Solution

Step 1: Recognize the series

Step 2: Rewrite the equation

Step 3: Solve for x∈(−π,π) x \in (-\pi, \pi)

Solution

Solution

Solution

Problem:

Given:

Now calculate:

Also,

✅ Therefore,

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

? GP and Function Relation

Solution

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Step 1: Write in powers of $a = 2^{1/6}$

Step 2: Use identity $a^6 = 2 \Rightarrow a^5 = \frac{2}{a}$

Step 3: Solve for $x \in (-\pi, \pi)$